Question

If the expression $[mx-1+\frac{1}{x}]$ is non-negative for all real $x$, then the minimum value of $m$ must be

• A. $-\frac{1}{2}$
• B. $0$
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $\frac{1}{4}$

$f\left(x\right)=mx-1+\frac{1}{x}f\left(x\right)=\frac{m{x}^2-x+1}{x}f\left(x\right)\ge 0$

for every $x\in R.\therefore {(-1)}^2-4m\le 04m\ge 1m\ge \frac{1}{4}$

$\therefore$ minimum value of $m$ is $\frac{1}{4}$.